Integrate x^(5/2) e^(-x): Solving w/ Substitution & √2π

In summary, the conversation is about how to integrate x^(5/2) e^(-x) dx from 0 to infinity, using the knowledge of the given integral \int_{-\infty}^{\infty}e^{-x^2/2} dx = \sqrt{2\pi}. One person suggests using Feynman's trick, while another suggests using partial integration. Ultimately, both methods will lead to the evaluation of the given integral.
  • #1
nikhilb1997
14
0

Homework Statement


Using [tex]\int_{-\infty}^{\infty}e^{-x^2/2} dx = \sqrt{2\pi}[/tex], Integrate x^(5/2) e^(-x) dx from 0 to infinty

2. The attempt at a solution

I tried substituting x = u^2/2 but i could not simplify further.
Please help me with the problem.
Thank you in advance.
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
You are supposed to use the knowledge of the result of the given integral to find out the integral you want. Changing the integral to one with ##-x## in the exponent instead of ##-x^2/2## is a good start. I also suggest looking up Feynman's trick of differentiating an integral with respect to a parameter (not the integration variable) and changing the order of integration and differentiation.
 
  • #3
Orodruin said:
You are supposed to use the knowledge of the result of the given integral to find out the integral you want. Changing the integral to one with ##-x## in the exponent instead of ##-x^2/2## is a good start. I also suggest looking up Feynman's trick of differentiating an integral with respect to a parameter (not the integration variable) and changing the order of integration and differentiation.
Thanks a lot feynman's trick was a good read. But if suppose i did have to use the given result then, what would be the method to go forward with?
 
  • #4
I would use Feynman's trick together with the given result. The alternative is doing partial integration. It all is going to boil down to evaluation of the given integral in the end.
 
  • #5
Orodruin said:
I would use Feynman's trick together with the given result. The alternative is doing partial integration. It all is going to boil down to evaluation of the given integral in the end.
I was just trying it i understood how it works. I should be able to use both together. Thanks a lot.
 

1. What is the purpose of using substitution when solving for the integral of x^(5/2) e^(-x)?

Substitution is used to simplify the integrand and make it easier to solve. In this case, substituting u = √x will eliminate the fractional exponent and make the integral easier to evaluate.

2. How do you choose the appropriate substitution for this integral?

The best substitution to use is one that will eliminate the complex or difficult components of the integrand. In this case, u = √x is a good choice because it eliminates the fractional exponent.

3. What is the process for solving this integral using substitution?

The first step is to choose the appropriate substitution, in this case u = √x. Then, substitute u into the integrand and solve for the new integral in terms of u. After integrating, substitute back in the original variable x to get the final solution.

4. What is the significance of the constant √2π in the solution?

The constant √2π is a result of the substitution and is known as the Jacobian. It is a factor that is used to account for the change in variables and is often present in integrals solved using substitution.

5. How do you know if your solution is correct?

You can check your solution by taking the derivative of the result and seeing if it matches the original integrand. If it does, then your solution is correct. You can also use a graphing calculator to graph the original integrand and your solution to visually confirm that they are the same.

Similar threads

  • Calculus and Beyond Homework Help
Replies
7
Views
936
  • Calculus and Beyond Homework Help
Replies
7
Views
710
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
11
Views
1K
  • Calculus and Beyond Homework Help
2
Replies
47
Views
2K
  • Calculus and Beyond Homework Help
Replies
4
Views
847
  • Calculus and Beyond Homework Help
2
Replies
44
Views
4K
Replies
5
Views
971
  • Calculus and Beyond Homework Help
Replies
22
Views
1K
Back
Top